Solve -2x/3x^2+4x+1+1/9x^2-1+1/x+1

$$\frac { - 2 x } { 3 x ^ { 2 } + 4 x + 1 } + \frac { 1 } { 9 x ^ { 2 } - 1 } + \frac { 1 } { x + 1 }$$

Evaluate

$$\frac{3x}{9x^{2}-1}$$

Short Solution Steps

Factor 3x^{2}+4x+1. Factor 9x^{2}-1.

$$\frac{-2x}{\left(x+1\right)\left(3x+1\right)}+\frac{1}{\left(3x-1\right)\left(3x+1\right)}+\frac{1}{x+1}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x+1\right)\left(3x+1\right) and \left(3x-1\right)\left(3x+1\right) is \left(3x-1\right)\left(x+1\right)\left(3x+1\right). Multiply \frac{-2x}{\left(x+1\right)\left(3x+1\right)} times \frac{3x-1}{3x-1}. Multiply \frac{1}{\left(3x-1\right)\left(3x+1\right)} times \frac{x+1}{x+1}.

$$\frac{-2x\left(3x-1\right)}{\left(3x-1\right)\left(x+1\right)\left(3x+1\right)}+\frac{x+1}{\left(3x-1\right)\left(x+1\right)\left(3x+1\right)}+\frac{1}{x+1}$$

Since \frac{-2x\left(3x-1\right)}{\left(3x-1\right)\left(x+1\right)\left(3x+1\right)} and \frac{x+1}{\left(3x-1\right)\left(x+1\right)\left(3x+1\right)} have the same denominator, add them by adding their numerators.

$$\frac{-2x\left(3x-1\right)+x+1}{\left(3x-1\right)\left(x+1\right)\left(3x+1\right)}+\frac{1}{x+1}$$

Do the multiplications in -2x\left(3x-1\right)+x+1.

$$\frac{-6x^{2}+2x+x+1}{\left(3x-1\right)\left(x+1\right)\left(3x+1\right)}+\frac{1}{x+1}$$

Combine like terms in -6x^{2}+2x+x+1.

$$\frac{-6x^{2}+3x+1}{\left(3x-1\right)\left(x+1\right)\left(3x+1\right)}+\frac{1}{x+1}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(3x-1\right)\left(x+1\right)\left(3x+1\right) and x+1 is \left(3x-1\right)\left(x+1\right)\left(3x+1\right). Multiply \frac{1}{x+1} times \frac{\left(3x-1\right)\left(3x+1\right)}{\left(3x-1\right)\left(3x+1\right)}.

$$\frac{-6x^{2}+3x+1}{\left(3x-1\right)\left(x+1\right)\left(3x+1\right)}+\frac{\left(3x-1\right)\left(3x+1\right)}{\left(3x-1\right)\left(x+1\right)\left(3x+1\right)}$$

Since \frac{-6x^{2}+3x+1}{\left(3x-1\right)\left(x+1\right)\left(3x+1\right)} and \frac{\left(3x-1\right)\left(3x+1\right)}{\left(3x-1\right)\left(x+1\right)\left(3x+1\right)} have the same denominator, add them by adding their numerators.

$$\frac{-6x^{2}+3x+1+\left(3x-1\right)\left(3x+1\right)}{\left(3x-1\right)\left(x+1\right)\left(3x+1\right)}$$

Do the multiplications in -6x^{2}+3x+1+\left(3x-1\right)\left(3x+1\right).

$$\frac{-6x^{2}+3x+1+9x^{2}+3x-3x-1}{\left(3x-1\right)\left(x+1\right)\left(3x+1\right)}$$

Combine like terms in -6x^{2}+3x+1+9x^{2}+3x-3x-1.

$$\frac{3x^{2}+3x}{\left(3x-1\right)\left(x+1\right)\left(3x+1\right)}$$

Factor the expressions that are not already factored in \frac{3x^{2}+3x}{\left(3x-1\right)\left(x+1\right)\left(3x+1\right)}.

$$\frac{3x\left(x+1\right)}{\left(3x-1\right)\left(x+1\right)\left(3x+1\right)}$$

Cancel out x+1 in both numerator and denominator.

$$\frac{3x}{\left(3x-1\right)\left(3x+1\right)}$$

Expand \left(3x-1\right)\left(3x+1\right).

$$\frac{3x}{9x^{2}-1}$$

Differentiate w.r.t. x

$$-\frac{3\left(9x^{2}+1\right)}{\left(9x^{2}-1\right)^{2}}$$

Steps Using Derivative Rule for Quotient

Factor 3x^{2}+4x+1. Factor 9x^{2}-1.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{-2x}{\left(x+1\right)\left(3x+1\right)}+\frac{1}{\left(3x-1\right)\left(3x+1\right)}+\frac{1}{x+1})$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x+1\right)\left(3x+1\right) and \left(3x-1\right)\left(3x+1\right) is \left(3x-1\right)\left(x+1\right)\left(3x+1\right). Multiply \frac{-2x}{\left(x+1\right)\left(3x+1\right)} times \frac{3x-1}{3x-1}. Multiply \frac{1}{\left(3x-1\right)\left(3x+1\right)} times \frac{x+1}{x+1}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{-2x\left(3x-1\right)}{\left(3x-1\right)\left(x+1\right)\left(3x+1\right)}+\frac{x+1}{\left(3x-1\right)\left(x+1\right)\left(3x+1\right)}+\frac{1}{x+1})$$

Since \frac{-2x\left(3x-1\right)}{\left(3x-1\right)\left(x+1\right)\left(3x+1\right)} and \frac{x+1}{\left(3x-1\right)\left(x+1\right)\left(3x+1\right)} have the same denominator, add them by adding their numerators.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{-2x\left(3x-1\right)+x+1}{\left(3x-1\right)\left(x+1\right)\left(3x+1\right)}+\frac{1}{x+1})$$

Do the multiplications in -2x\left(3x-1\right)+x+1.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{-6x^{2}+2x+x+1}{\left(3x-1\right)\left(x+1\right)\left(3x+1\right)}+\frac{1}{x+1})$$

Combine like terms in -6x^{2}+2x+x+1.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{-6x^{2}+3x+1}{\left(3x-1\right)\left(x+1\right)\left(3x+1\right)}+\frac{1}{x+1})$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(3x-1\right)\left(x+1\right)\left(3x+1\right) and x+1 is \left(3x-1\right)\left(x+1\right)\left(3x+1\right). Multiply \frac{1}{x+1} times \frac{\left(3x-1\right)\left(3x+1\right)}{\left(3x-1\right)\left(3x+1\right)}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{-6x^{2}+3x+1}{\left(3x-1\right)\left(x+1\right)\left(3x+1\right)}+\frac{\left(3x-1\right)\left(3x+1\right)}{\left(3x-1\right)\left(x+1\right)\left(3x+1\right)})$$

Since \frac{-6x^{2}+3x+1}{\left(3x-1\right)\left(x+1\right)\left(3x+1\right)} and \frac{\left(3x-1\right)\left(3x+1\right)}{\left(3x-1\right)\left(x+1\right)\left(3x+1\right)} have the same denominator, add them by adding their numerators.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{-6x^{2}+3x+1+\left(3x-1\right)\left(3x+1\right)}{\left(3x-1\right)\left(x+1\right)\left(3x+1\right)})$$

Do the multiplications in -6x^{2}+3x+1+\left(3x-1\right)\left(3x+1\right).

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{-6x^{2}+3x+1+9x^{2}+3x-3x-1}{\left(3x-1\right)\left(x+1\right)\left(3x+1\right)})$$

Combine like terms in -6x^{2}+3x+1+9x^{2}+3x-3x-1.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{3x^{2}+3x}{\left(3x-1\right)\left(x+1\right)\left(3x+1\right)})$$

Factor the expressions that are not already factored in \frac{3x^{2}+3x}{\left(3x-1\right)\left(x+1\right)\left(3x+1\right)}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{3x\left(x+1\right)}{\left(3x-1\right)\left(x+1\right)\left(3x+1\right)})$$

Cancel out x+1 in both numerator and denominator.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{3x}{\left(3x-1\right)\left(3x+1\right)})$$

Consider \left(3x-1\right)\left(3x+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{3x}{\left(3x\right)^{2}-1})$$

Expand \left(3x\right)^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{3x}{3^{2}x^{2}-1})$$

Calculate 3 to the power of 2 and get 9.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{3x}{9x^{2}-1})$$

For any two differentiable functions, the derivative of the quotient of two functions is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the denominator squared.

$$\frac{\left(9x^{2}-1\right)\frac{\mathrm{d}}{\mathrm{d}x}(3x^{1})-3x^{1}\frac{\mathrm{d}}{\mathrm{d}x}(9x^{2}-1)}{\left(9x^{2}-1\right)^{2}}$$

The derivative of a polynomial is the sum of the derivatives of its terms. The derivative of a constant term is 0. The derivative of ax^{n} is nax^{n-1}.

$$\frac{\left(9x^{2}-1\right)\times 3x^{1-1}-3x^{1}\times 2\times 9x^{2-1}}{\left(9x^{2}-1\right)^{2}}$$

Do the arithmetic.

$$\frac{\left(9x^{2}-1\right)\times 3x^{0}-3x^{1}\times 18x^{1}}{\left(9x^{2}-1\right)^{2}}$$

Expand using distributive property.

$$\frac{9x^{2}\times 3x^{0}-3x^{0}-3x^{1}\times 18x^{1}}{\left(9x^{2}-1\right)^{2}}$$

To multiply powers of the same base, add their exponents.

$$\frac{9\times 3x^{2}-3x^{0}-3\times 18x^{1+1}}{\left(9x^{2}-1\right)^{2}}$$

Do the arithmetic.

$$\frac{27x^{2}-3x^{0}-54x^{2}}{\left(9x^{2}-1\right)^{2}}$$

Combine like terms.

$$\frac{\left(27-54\right)x^{2}-3x^{0}}{\left(9x^{2}-1\right)^{2}}$$

Subtract 54 from 27.

$$\frac{-27x^{2}-3x^{0}}{\left(9x^{2}-1\right)^{2}}$$

Factor out 3.

$$\frac{3\left(-9x^{2}-x^{0}\right)}{\left(9x^{2}-1\right)^{2}}$$

For any term t except 0, t^{0}=1.

$$\frac{3\left(-9x^{2}-1\right)}{\left(9x^{2}-1\right)^{2}}$$